 Okay so let us continue with our discussion uhhh so you know the general theme is a theme of uhhh zeros of analytic functions and uhhh what we saw last time was the so called open mapping theorem which said that if you take a non constant analytic function uhhh defined on a domain then it is an open map so it maps open sets to open sets the image of an open set under such a map is again an open set okay uhhh. So what I am going to do now is uhhh uhhh prove the so called uhhh you know the so called inverse function theorem which is uhhh uhhh which is essentially the statement that you know if you have an analytic function that does not van that uhhh whose derivative does not vanish at a given point then there is a small disc around that point uhhh there is a small neighbourhood around that point where the analytic function is one to one okay and not only that uhhh the inverse function which can be defined because it is one to one is also analytic okay and the inverse function has is given by an integral formula okay. So this is what we are going to see today so let me let me let me uhhh uhhh let me put that as a title of today's uhhh talk or rather the aim of today's talk uhhh the uhhh inverse function theorem. So uhhh so what is the uhhh what is the uhhh uhhh uhhh theorem uhhh and well I will call this a simple statement. So the simple statement is uhhh let uhhh uhhh in every neighbourhood in uhhh at every at every for every point for every point for a point uhhh z0 uhhh where the analytic function f of z has non vanishing derivative okay which is the statement that f dash of z0 uhhh uhhh uhhh is non-zero okay uhhh there there is uhhh there is uhhh a neighbourhood of z0 restricted to which which uhhh f is one to one or or or injective and uhhh the the the function the the the the inverse function f inverse uhhh that is that is so defined on uhhh the uhhh image of this neighbourhood is also analytic. So this is the statement of the the inverse function theorem the simple statement okay. So if I so if I draw a diagram it is something like this. So here is a so there is some domain uhhh uhhh some domain I am drawing a bounded domain but bounded domain enclosed by a uhhh simple closed curve but it may not look like that it is some domain and by by a domain we always mean an open connected set okay and uhhh this is in the source complex plane which is the z plane and you see on this domain D uhhh there is defined analytic function w equal to f of z uhhh which takes values in the w plane or the omega plane which is another copy of complex numbers. So this is a w plane and uhhh I am considering a point suppose I consider a point z0 such that uhhh the corresponding point w0 which is equal to f of z0 uhhh is its image okay and suppose I assume that f dash of z0 is not equal to 0. So that means that uhhh this is the fact uhhh that I have put here in the hypothesis that it has non vanishing derivative okay f dash of z is the derivative of f of z and its value adds z0 uhhh so when I say non uhhh uhhh the non vanishing derivative is adds z0 okay. Then uhhh the the claim is that that I can find a neighbourhood for example a small disc surrounding z0 okay where restricted to which the function becomes 1 to 1 okay and uhhh if I call this uhhh discuss uhhh uhhh something let me call this as D0 okay uhhh uhhh r let me even call it delta not. Then what happens is that f of delta not uhhh uhhh will be uhhh uhhh uhhh uhhh will be a will will be a neighbourhood of w0 in fact it will be a disc you can uhhh very well expected to be a disc like neighbourhood of w0. So uhhh so let me say you know uhhh okay so I need to move this and draw something like this here uhhh w0 is so this is f of delta not is this disc disc like neighbourhood and uhhh this is uhhh uhhh w0 is just f of z0 and what happens is that from this image disc from this image f of delta delta not you have a an inverse function which is given by f inverse it is given by w uhhh z equal to f inverse of w that is inverse function okay. And the fact is that uhhh the fact is first of all that f restricted to this uhhh this disc delta not is 1 to 1 okay. There is a disc delta not a neighbourhood of z0 restricted to which f is 1 to 1 and therefore and the image of this delta not will be the set f of delta not you must remember that uhhh if you recall the uhhh open mapping theorem because f is a of course uhhh uhhh mind you it has now uhhh uhhh the derivative at z0 does not vanish therefore f cannot be a constant function. So, it is a non-constant analytic function because if the function is a constant analytic function then this derivative will be 0 identically 0. So, the fact that the analytic function has non vanishing derivative at one point tells you that uhhh the function is a non-constant analytic function. And what we saw in the previous lecture was that such a non-constant analytic function is an open mapping that is it always takes open sets to open sets. So, if you take this disc delta not which is this open disc centered at z0 of suitable radius okay. Then its image will always be uhhh an open set okay which I am drawing for convenience like a disc okay and the theorem the inverse function theorem says that f restricted to delta not is injective it is 1 to 1. Therefore, you can f inverse is defined for every point w in f of delta not. So, there is a inverse function like this set theoretic inverse function that is because for every z there is a unique w uhhh for every z uhhh you have w equal to f of z and for every w here there is a unique z here such that f of z equal to that w okay. So, you have a inverse set theoretic inverse map and what the inverse function theorem says is that this is not this inverse function is actually analytic okay. And in other words what you are saying is f restricted to delta not is a holomorphic or analytic isomorphism of delta not with f of delta not it is a holomorphic isomorphism analytic isomorphism. Because this way the map is uhhh holomorphic or analytic and f inverse is also holomorphic and analytic and both are inverses of each other because f is 1 to 1 okay. So, this is the inverse function theorem okay this is a simple statement okay. So, the more complicated uhhh statement is uhhh is giving you is this is the statement that will give you what this delta not exactly is it will give you what this delta not is it will tell you how to calculate this it will give you a formula for f inverse okay that is the more deeper statement. So, let me write down the more deeper statement here okay. So, theorem uhhh this is uhhh uhhh more uhhh I should say more uhhh uhhh involved statement what is the more involved statement. So, uhhh let uhhh uhhh f be analytic uhhh at a point z not okay mind you this means that f is uhhh analytic function defined in an open neighbourhood containing z not. So, it is analytic at a analytic at a point of course means analytic and neighbourhood of the point that is what analytic means okay differentiable in a neighbourhood of a point which is analytic in the neighbourhood point uhhh. Suppose uhhh f dash of z not is not equal to z okay. Suppose the derivative of f does not vanish at z not of course it is analytic therefore that is holomorphic. So, it is it means it is differentiable and you know it is differentiable once and it is differentiable infinitely many times because that is the specialty of analytic functions. So, first derivative exists and higher derivative exists the fact is the first derivative at z not is non zero okay that is the assumption. Choose uhhh uhhh row uhhh so, so the uhhh so let me rub this off and say something I want to say that since you have a non so, so this implies that f is a non constant analytic function this is this this is something that I should uhhh so, f is a non constant analytic function on uhhh uhhh uhhh uhhh uhhh in a neighbourhood uhhh uhhh around z not okay. So, it is a non constant analytic function of course when I say around z not I mean in uhhh disc surrounding z not okay and I you know the if I if a function is a non constant analytic function you know that it is uhhh zeros are isolated okay a non constant analytic function for a non constant analytic function zeros are isolated and uhhh mind you uhhh if I if you take w not to be the value of f at z not then z not is a 0 of f of z minus w not okay this is a trick that we have always been using uhhh all these uhhh during all the previous lectures okay. So, uhhh put w not equal to f of z not then uhhh uhhh uhhh z not is a 0 of uhhh order 1 of f of z minus w not okay. So, let me explain this of course f if you take the function f of z minus w not okay and if I plug in z equal to z not I will get 0. So, z not is z equal to z not is a 0 of this function no doubt okay and the fact that uhhh the derivative of this function at z not is the same as f dash of z not because minus w not is only a constant. If I take the derivative of this I will simply get f dash z and if I plug in z not I will get f dash z not this is not 0 since the first derivative is not 0 it means it is 0 of order 1 okay. You see uhhh if you want uhhh maybe I can uhhh maybe I can recollect uhhh something from uhhh say basic complex analysis uhhh for you. You see uhhh you can recall uhhh uhhh an analytic uhhh an an analytic function function uhhh non-constant of course non-constant uhhh has a 0. So, I call the function g of z okay has a 0 of uhhh of uhhh order uhhh r at z equal to z not if you know uhhh g of z can be written as z minus z not power r times h of z okay with you know uhhh with h of z analytic uhhh around z not at z not and h dash of z not is not equal to 0. This is what is meant by uhhh a function having a 0 of certain order at the point z minus z not. That means you see a function having a 0 of order r at z equal to z not means z minus z not to the power of r can be factored out from the uhhh Taylor series of the function centered at the point z not that is exactly what it says okay. Of course if I put if I put z equal to z not here I will get 0 okay and the uhhh the the order of the 0 is the is this power is the power of z minus z not. So, you know what you must understand is that uhhh it follows it follows that uhhh you know if you calculate g dash of z not and so on and you go on up to g uhhh you know r minus 1 of z not they will always be 0 h should not have a 0 at z not that is correct that is correct it is not h dash thanks yeah that is uhhh see the 0 when you write the function in this form z not should not be a 0 of this part that is important that is correct that is correct thank you for pointing out I missed that. So, you see so h of z not is not it is not h dash you are right h of z not is not 0 okay and see what happens is that if you start differentiating g uhhh up to r minus 1 times and you substitute z not you will still get 0 okay but if you differentiate it r times you want get 0 okay because if you differentiate it r times uhhh if you use a product rule and do it okay then you will have one term uhhh which will involve uhhh uhhh so you know uhhh so for example you know if if if so you know uhhh so for example uhhh if you take g of z let us let us put z minus z not power let us put 1 into h of z suppose r equal to 1 you know if you differentiate it once you will get uhhh you know by product rule z minus z not constant I will get h dash of z plus h of z into differentiation of z minus z not is just 1 and if I calculate g dash of z not I will simply get h of z not and that is not 0. So if r is equal to 1 then uhhh h of z not is uhhh non zero and g dash of z not is non zero okay and so you can now you can you can generalize it okay uhhh if you have this situation then g r of z not will not be 0 the so you know g1 g dash is the first derivative g I will put a round bracket here saying that this is the r minus 1 derivative but g uhhh if I calculate g rth derivative at z not that would not be 0 okay. So so so this is so this is example when r equal to 1 alright similarly you you can try it out for r equal to 2 3 and so on alright. So the point is that when you uhhh you can write g as z minus z not to that power which is the order of the 0 times a function which does not have a 0 at z not okay and what does this mean see this is just a reflection of the fact that the Taylor series for g centered at z not starts only from uhhh the the Taylor coefficients are all 0 uhhh up to uhhh the first the first r coefficients are all 0 that is what it means see because you know uhhh so you know if if you if you look at it in another way you see what is g of z as a Taylor series it will be uhhh a not plus a1 z minus z not into uhhh plus a2 z minus z not squared and so on okay and what are the ai the ai's are just uhhh the uhhh ith derivative of g at z not by uhhh so a n is just the nth derivative of g at z not by factorial n right these are this is the Taylor series alright and uhhh if you so you know the the the derivatives from the first uhhh uhhh and of course g of z not is also 0 mind you so a not is 0 okay a1 is a multiple of the first derivative which is 0 and so on all the first up to the first r minus 1 a a not up to a r minus 1 will all be 0 so the power series will start with the first term which will correspond to z minus z not to the power of r. So what will happen is that g of z will just be z minus z not power r it will start only like this this will be the first term I mean this will be the first power that you will get and of course uhhh the the the coefficient will be of course a r then you will get then it will start from here it will be a r plus 1 uhhh z minus z not to the power of r plus 1 and it will go on like this and the fact is that if you factor out z minus z not power r from this what you will get out what you will get the the remaining that you will get will be the power series expansion for h of z and centered at z not and the power the power series have to be the same because there is an identity theorem for power series. The power series uhhh surround uhhh expansion at a point the Taylor series is unique okay and it is unique basically because it is given by these derivatives and the derivatives are of course unique okay. So what you must understand is this is exactly what is happening when you take a 0 of order r if z not is a 0 of order r then you know uhhh the that is signified by saying that the first r minus 1 derivatives uhhh including the function which is sort of a the 0 derivative okay they all vanish at that point but the rth derivative will not vanish. So in particular you know if you take r equal to 1 that is the situation here in this case z not is a 0 of f z minus w not okay and z not is only a 0 of order 1 because it is the derivative of this function is f dash of z and when you evaluated z not you will get f dash of z not and f dash of z not is not 0 okay. If it had been a 0 of order greater than 1 then f dash of z not would have also become 0 okay. So that should tell you that this is a derivative of I mean it is a 0 of order 1 for this function okay now uhhh uhhh we can find a delta uhhh okay so uhhh so let me say something else so let me go back. So this is about I mean this is all this was just to recall uhhh what order means okay. Now go back to this function see f of z is a non constant analytic function. The f of z minus w not is also a non constant analytic function and the zeros of a non constant analytic function are isolated all the zeros are isolated. Therefore the the the zeros of this non constant analytic function are also isolated in particular the 0 z not is also isolated. So there is a disc surrounding z not where there are no other zeros okay. So so since the zeros of a non constant analytic function are isolated there exists a row positive uhhh such that f of z minus minus w not has no 0 in 0 strictly less than mod z minus z not less than or equal to rho okay. This is because of isolation of uhhh zeros of a non constant analytic function okay. So that we in other words I have excluded 0 mind you because when I put 0 z has to be equal to z not and z not is of course is 0. But this is a deleted closed disc centred at z not on which there are no other zeros and that is such a disc exists because of uhhh you know uhhh isolation of zeros of a non constant analytic function okay. So so there is a row like this alright. Now also there exists a delta positive such that the modulus of f of z minus uhhh w not is greater than or equal to delta on for z with mod z minus z not equal to rho okay. I mean this is the fact that we have used uhhh couple of times even in the even in the uhhh earlier proofs for example in the proof of uhhh uhhh the open mapping theorem alright. So uhhh so what I will have to do is uhhh I wrote it here. So let me rub this off so that I can continue with the statement of the theorem okay. So uhhh so let me continue here uhhh so you see the uhhh uhhh this is something that we have uhhh you know always been using the function f of z minus w not uhhh is uhhh uhhh if you take that function that function is not going to vanish on the on this deleted neighbourhood. So in particular if you take the boundary of this deleted neighbourhood which is the circle mod z minus z not equal to rho circle centered at z not radius rho it is not going to vanish on that circle okay. So it is uhhh but that circle is both you know it is compact connected and uhhh this is mod f z minus w not is a continuous function the continuous real valued function on a compact uhhh connected set therefore its image on the real line is going to be a compact connected subset of the real line and therefore it has to be a closed interval okay and uhhh delta is the minimum value uhhh or in that closed interval namely it is the left end point of the closed interval okay. So such a delta can be found okay the fact is that so what the inverse function theorem says is uhhh uhhh for each uhhh for each uhhh z in mod z minus z not lesser than uhhh lesser than uhhh rho okay uhhh f of z is uhhh uhhh f dash of z is not equal to 0 okay. So see the derivative uhhh you assume is uhhh it does not vanish at z not okay but the fact is that in this disc okay the derivative is never going to vanish at any point that is once that is one claim okay uhhh f is 1 1 1 to 1 on mod z minus z not 6 strictly less than rho this is the disc on which f is 1 to 1 okay uhhh for every for every w with mod w minus w not strictly less than delta okay this is the uhhh this is the target region uhhh in the uhhh w plane okay there exists a unique z with mod z minus z not strictly less than rho such that uhhh f of z is w okay and in fact uhhh z is equal to f inverse w is 1 by 2 pi i integral over mod z mod zeta minus z not equal to rho okay uhhh uhhh zeta uhhh f dash of zeta d zeta by f of zeta minus w this is the inverse this is the inverse formula for the inverse function. So you see this expresses f inverse of w okay uhhh of course this is for mod w minus w not lesser than delta okay. So the the formula for the inverse function is given in this is given like this okay the formula for the inverse function is given like this uhhh and the claim is that f inverse as a function of w is also analytic further of w for uhhh mod w minus w not less than so this is the uhhh this is the strong uhhh or the more involved statement of the inverse function. So it tells you what is the disc it is a certain disc centred at z not radius rho okay and it also tells you what is the uhhh target disc the target uhhh region on which you can write out the formula for the inverse that is also a disc centred at w not and it is radius this is delta where delta is the uhhh you know uhhh minimum value of f of z of of mod f z minus w not on uhhh for z varying on this boundary disc okay this is what the inverse function is okay. So uhhh so let us let us try to uhhh prove this. So the first thing I want to say is that uhhh uhhh you know uhhh uhhh the the first question that I want to think about is the following uhhh so so see there are several things I have to prove I have to show it is one to one okay then I have to show that uhhh uhhh once once it is one to one then I have to then I will have to show that the inverse function is given by this formula then I have to show that f inverse is an analytic function of w okay these are the three things I have to do. Now how do I show it is one to one I actually go back to the essentially to the proof of the open mapping theorem okay. So it takes me back to the proof of the open mapping theorem and what is the proof. So you see so you know let us let us look at this see what is uhhh number of zeros of f of z minus w not in mod z minus z not equal to o okay. So it is you know the basically it is the argument principle it is the counting principle which is just uhhh counting the number of zeros of an analytic function inside a closed simple closed curve which is just given by the argument principle which is just the residue theorem applied to the logarithmic derivative of the function. So what is the number of zeros if I call that as n of w not the number of zeros uhhh of uhhh f of z minus w not in this disk okay this is this is given by what I mean it is given by well uhhh by the by the residue theorem it has I mean by the argument principle which is residue theorem applied to the logarithmic derivative of this it is just 1 by 2 pi i integral over mod z minus z not is equal to rho okay of d log uhhh f of z minus w not this is what is if you take d log of a function of an analytic function and integrate it over a uhhh boundary curve simple closed curve which is smooth contour and divide by 2 pi i you will get simply the uhhh number of zeros minus number of poles okay. But of course if the function has no uhhh if it has no poles then you will just be getting the number of zeros counted with multiplicity and in this case uhhh uhhh uhhh what is this this integral this integral is just 1 by 2 pi i integral over mod z minus z not is equal to rho d log of that is just derivative of that logarithmic derivative is the derivative of this by uhhh by this function okay. So what you should understand is uhhh when I do whenever you do an integration the variable of integration lies on the boundary curve okay. So for z lying on the boundary curve you know that mod of f of z minus w not does not vanish because you what you have assumed is here what you have assumed here is that for z lying on the boundary curve mod z minus z not equal to rho the mod of f of z minus w not is greater than or equal to delta which is a positive number okay and mind you delta is positive because this does not vanish and why why this does not vanish is because it does not vanish the only point where it vanishes in this in this close disc center at z not radius rho is at the center okay and the boundary it certainly does not vanish. So this this quantity in the denominator uhhh that never has a zero okay uhhh this this quantity it does not have a zero on the boundary it has a zero inside and that is what is counted by this and this and this is equal to 1 of course because the number of zeros uhhh number of uhhh the number of zeros of this is just the number of times uhhh f of z takes the value w not and that is only once at z equal to z not that is our assumption because it has only one zero and that zero is of order one okay so it is counted only once if zero is of order m you have to count it as m zeros okay but this is zero of order one so it is counted only as one zero and only at one point so you get one okay this is what you get but now more generally if you recall in the open mapping theorem the proof of the open mapping theorem what we did was we also defined n w the number of times uhhh the function f of z uhhh uhhh f of z takes the value w and how was that defined that was defined as 1 by 2 pi i integral over mod z minus z not equal to rho of d log f of z minus w this is how we defined it and what is this uhhh how was this defined I mean which is which is what is uhhh which is which is which is equal to uhhh the following uhhh this is f dash of z dz by f of z minus w okay and for for what values of w does this integral make sense this is for uhhh uhhh w with mod w minus w not uhhh strictly less than delta strictly less than delta you see the fact is uhhh the the way we have chosen uhhh delta is such that the for on the boundary curve the distance of f of z from w not is greater than or equal to delta okay so if you choose a w whose distance from w not is less than delta then the distance of f of z from that such a w can never be 0. So this quantity does not vanish on the boundary curve therefore this integral is well defined and this integral will give you the number of times the analytic function f of z assumes the value w in uhhh inside this uhhh curve inside this circle centre at z not radius rho okay and in fact what we proved last time is if you uhhh if you if you recall we showed uhhh in the proof of uhhh and the course of the proof of the open mapping theorem that uhhh n of w is an analytic function of w it is an analytic function of w n of w is an analytic function of w right that is something that we proved and then what we concluded was you see n of w is an analytic function of w it is defined on this disc uhhh centred at w not uhhh radius delta and uhhh but the point is that n of w takes only integer values because it counts the number of zeros okay it counts the number of times it counts the number of zeros of f of z minus w which is the same as counting the number of times uhhh f takes the value w for uhhh for z inside the disc centred at z not radius rho okay and that is an integer. So you have an analytic function of uhhh whose values are in the integers okay such a function has to be a constant because n of w is an analytic function so it is continuous analytic function is always continuous. So the image of this function image of this disc which is a connected set under a continuous map is always a connected set. So what you should get is that the image of n of w is a connected subset of the integers that has to be a single integer. So what this will tell you uhhh but n of w being integer value forces n of w is equal to constant it is equal to n of w not which is equal to 1 because n of w not is 1 because the number of times f assumes the value w not is precisely once that is how the disc has been chosen because the disc has been chosen to have only this only 1 0 the 0 z not of f minus w not okay. So but what does this tell you this tells you that for every w in uhhh such that mod w minus w not is less than delta the number of times f assumes that value w is exactly 1 okay. So what this will tell you is it will tell you 2 things it will tell you that the image of uhhh the disc mod z minus z not strictly less than rho contains this disc okay every value w which is within a distance of delta from w not is assumed by f at a point z which lies inside the disc mod z minus z not uhhh uhhh strictly less than rho. So it tells you 2 things it tells you that f of mod z minus z not less than rho contains the set mod mod w minus w not strictly less than delta it tells you this plus it also tells you that the fact that it assumes every value once is also telling that it is 1 to 1 f is 1 to 1 you had it see the uhhh what you must understand is the uhhh the fact that it assumes every value uhhh f f f of z minus f of z assumes the value w at a for only 1 point z okay and not for more than 1 point z but that is exactly saying that f is 1 to 1. So so so we get this is this and uhhh f is 1 to 1 on this disc okay. So we have proved this fact we have proved the fact that f is 1 to 1 on the whole disc and it takes all the values uhhh in the image disc it takes each value in the image disc exactly one time okay we have that proves this right and uhhh so that is one thing uhhh the okay the uhhh so so that proves uhhh uhhh so that proves if you go back to the statement of the theorem we have proved that f is 1 to 1 on this disc we have proved that for every w with mod w minus w not less than delta that the disc in the target uhhh complex plane w plane okay there exists a unique z that there exists a unique z is the 1 to 1 nature of f and that there exists an z is because n of w is equal to 1 the fact that n of w equal to 1 means that there is a point z for which f of z takes the value w and there is only 1 point okay. So there exists a unique z with mod z minus z not less than rho such that f of z equal to w okay. So we have proved that so what is left out is only to show that the formula for f inverse in terms of w is given by this expression okay this is the only thing that is left out I have to show that this formula holds and I have to show that f inverse is also an analytic function so this is this is what is left out. So I will I will I will do that in the in the next lecture okay.